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It intercepts the y axis at (0, 5) and it intercepts the x axis at (-2.3, 0) passing through the I, II and III quadrants
6x - 3y = 12y intercept at x = 0 is y = 4x intercept at y = 0 is x = 2the graph is straight line passing through the two points: (0, 4) and (2, 0)
Restate the question: How do you graph 7x+2y=14? The easiest way to graph this equation is to find the intercepts. . . To find the y-intercept, let x=0: 7(0)+2y=14 <=> 2y=14 <=> 2y/2=14/2 <=> y=7, so mark the point (0,7) on the graph. To find the x-intercept, let y=0: 7x+2(0)=14 <=> 7x=14 <=> x=2, so mark (2,0) on the graph and draw a straight line through the 2 points. With a little practice, and if the numbers are 'nice', you can almost do this by just looking at the equation, or you can think of (0,_),(_,0). My students call this "googly eyes"!
If D > 0 then the graph intersects the x-axis 2 times.If D = 0 then the the x-axis is tangent to the graph.If D < 0 then the graph doe not intersects the x-axis.
Two ways. You can solve for y and then use slope intercept or use x and y intercepts. If x=0 then y=5 (0,5) If y=0 then x=-6 (-6,0) Graph those two points and then draw a line through them.